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I know there are lots of related questions on here, but I can't seem to find what I'm looking for.

Given some event with, say, a $1$ in $1{,}000{,}000$ probability (e.g., $7$ being chosen randomly as a number between $1$ and $1{,}000{,}000$), I'd like to get a rough idea of the probability of seeing that event happen at least once in $1{,}000{,}000$ trials. (The temptation is somehow to say that you get roughly even odds.)

I understand that the probability of seeing a $1$-in-$n$ event occur at least once in $n$ trials is simply $1 - \frac{(n-1)^n}{n^n}$, but when $n$ is large, computing such values is difficult (for me at least).

Hence I'd like an idea as to whether this converges as $n$ approaches infinity.

Just from playing around with some values up to ($n=144$), it seems that the value converges towards $\sim0.633$.

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So my questions relate to understanding this more.

  • Does this value indeed converge?
  • If so, does the resulting value have any significance?
  • Conversely, for an event with a 1-in-$n$ chance, is there a way to characterise (in the general case) how many trials would be needed to see such an event at least once with $p\approx 0.5$?
badroit
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  • After rooting around a bit more with the answer in hand, turns out this question was a duplicate: http://math.stackexchange.com/questions/6140/help-with-a-specific-limit-left-dfracn-1n-rightn-as-n-rightarrow. – badroit Nov 10 '13 at 02:13

1 Answers1

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We have $$\lim_{n\to\infty} \left(\frac{n-1}{n}\right)^n=e^{-1}.$$ The number $e$, the base of natural logarithms, is of great importance.

When $n$ is largish (and it can be much smaller than $10^6$), the number $\left(\frac{n-1}{n}\right)^n$ is very close to $e^{-1}$.

For the last question, the probability of at least one event in $k$ trials is $$1-\left(1-\frac{1}{n}\right)^k.$$ We want this to be $\frac{1}{2}$, so we want to solve the equation $$\left(1-\frac{1}{n}\right)^k=\frac{1}{2}.$$ To solve, take the logarithm of both sides.

André Nicolas
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  • $1-\left(\frac{k-1}{k}\right)^k$ strictly decreases from $1$ to $1-\frac1e$ so it can't ever be $\frac12$. – Geoff Pointer Jan 14 '14 at 23:43
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    @GeoffPointer: OP's third question is about probability of individual success $1/n$, with not necessarily matching number of trials, and asks for the number of trials. – André Nicolas Jan 15 '14 at 14:36
  • For some values of $n$ and $k$, you will find $\left(1-\frac{1}{n}\right)^k\approx e^{-k/n}$ – Henry Jan 20 '24 at 09:55